How do you find the constant of integration for intf'(x)dx if f(2)=1?

Aug 19, 2014

Normally, we want this integral function to be specified with a capital $f$, so that we can specify the antiderivative as $f \left(x\right)$.

However, using your variable naming, let's say that $F \left(x\right)$ is the antiderivative of $f ' \left(x\right)$, then by the Net Change Theorem, we have:

$f \left(x\right) = F \left(x\right) + C$

Therefore, the constant of integration is:

$C = f \left(x\right) - F \left(x\right)$
$= f \left(2\right) - F \left(2\right)$
$= 1 - F \left(2\right)$

This is a simple answer, however for many students, it is very difficult to this this abstractly. So, let's look at a concrete example:

$F \left(x\right) = {x}^{3}$ to match your variables
$F ' \left(x\right) = f ' \left(x\right) = 3 {x}^{2}$ to match your variables
$f \left(x\right) = \int 3 {x}^{2} \mathrm{dx}$
$= {x}^{3} + C$
$= F \left(x\right) + C$

Now, substitute the given values:

$f \left(2\right) = {x}^{3} + C = 1$
${2}^{3} + C = 1$
$F \left(2\right) + C = 1$
$C = 1 - F \left(2\right)$

So, if an abstract problem makes it difficult for you to find a solution, start with a concrete one to help you find a pattern.